#### Question

Prove that:

\[\sec\theta + \tan\theta = \frac{\cos\theta}{1 - \sin\theta}\]

#### Solution

\[\sec\theta + \tan\theta\]

\[ = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\]

\[ = \frac{1 + \sin\theta}{\cos\theta}\]

\[ = \frac{\cos\theta\left( 1 + \sin\theta \right)}{\cos^2 \theta}\]

\[ = \frac{\cos\theta\left( 1 + \sin\theta \right)}{1 - \sin^2 \theta} \left( \sin^2 \theta + \cos^2 \theta = 1 \right)\]

\[= \frac{\cos\theta\left( 1 + \sin\theta \right)}{\left( 1 + \sin\theta \right)\left( 1 - \sin\theta \right)}\]

\[ = \frac{\cos\theta}{1 - \sin\theta}\]

Is there an error in this question or solution?

Solution Prave That: Sec θ + Tan θ = Cos θ 1 − Sin θ Concept: Application of Trigonometry.